Find or evaluate the integral by completing the square. \[ \int_{0}^{1} \frac{d x}{x^{2}-2 x+2} \]

To evaluate the integral \[ \int_{0}^{1} \frac{d x}{x^{2}-2 x+2}, \] we start by completing the square in the denominator. The expression \(x^2 - 2x + 2\) can be rewritten as follows: \[ x^2 - 2x + 2 = (x - 1)^2 + 1. \] This transforms our integral into \[ \int_{0}^{1} \frac{d x}{(x - 1)^2 + 1}. \] Now, we recognize that \(\frac{1}{(x - 1)^2 + 1}\) resembles the standard form for the inverse tangent function. Specifically, we use the formula: \[ \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C. \] Here, \(a = 1\) and \(x\) is replaced by \((x - 1)\), which gives us: \[ \int \frac{d x}{(x - 1)^2 + 1} = \tan^{-1}(x - 1) + C. \] Now, we can evaluate the definite integral: \[ \int_{0}^{1} \frac{d x}{(x - 1)^2 + 1} = \left[ \tan^{-1}(x - 1) \right]_{0}^{1}. \] Calculating the limits, we have: 1. At \(x = 1\): \(\tan^{-1}(1 - 1) = \tan^{-1}(0) = 0.\) 2. At \(x = 0\): \(\tan^{-1}(0 - 1) = \tan^{-1}(-1) = -\frac{\pi}{4}.\) Thus, we can substitute these values into our integral: \[ \int_{0}^{1} \frac{d x}{(x - 1)^2 + 1} = 0 - \left(-\frac{\pi}{4}\right) = \frac{\pi}{4}. \] Therefore, the final result of the integral is \[ \int_{0}^{1} \frac{d x}{x^{2}-2 x+2} = \frac{\pi}{4}. \]